--- title: "Interfacing 1408 DAC for 0 to 10 V" subject: "Instrumentation II" chapter: 4 key: "semester_5/instrumentation_ii:dac1408-0-10v" source: https://www.bctengineer.com/q/semester_5/instrumentation_ii/dac1408-0-10v site: BCT Engineering --- # Interfacing 1408 DAC for 0 to 10 V Instrumentation II · Chapter 4: Interfacing A/D And D/A Converters - Asked in: 2076 Chaitra, 8 marks - Video: [DAC interfacing with 8086 | ALP program | | MPMC | Lec-47 | Bhanu Priya](https://www.youtube.com/watch?v=3XjjUs8Cm7M) — Education 4u - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/instrumentation_ii/dac1408-0-10v - Whole chapter: https://www.bctengineer.com/md/semester_5/instrumentation_ii/chapter-4 ## Question Design the interfacing of 1408 DAC with an output port of address AF H for 0v to 10v range. Note that take appropriate values for resistors and capacitors. ## Short answer DAC1408 sinks current; Iref = 2 mA from +10V via 5 kΩ, feedback resistor 5 kΩ yields Vout = 0–9.96 V. Address AFH is decoded, and data is latched by a 74LS373. A 0.1 µF compensation capacitor is placed from pin 16 to VEE. ## Answer #### DAC1408 (MC1408) basics The MC1408 is an 8-bit multiplying DAC with a current output. Its full-scale output current is Io = Iref x 255/256, where Iref = Vref / R14. Io is a current **sink** into pin 4, so the op-amp (inverting input at pin 4) supplies this current through Rf. The op-amp output is therefore positive: Vout = +Io x Rf. #### Design values for 0-10 V - Vref(+) = +10 V, R14 (pin 14) = 5 kohm, so Iref = 10 V / 5 kohm = 2 mA. Pin 15 (Vref-) is connected to ground through R15 = 5 kohm (equal to R14). - Vout = Io x Rf, with Rf = 5 kohm: Vout = 2 mA x 5 kohm x N/256 = 10 V x N/256. For N = FFH, Vout = 9.96 V; for N = 00H, Vout = 0 V. - Supply: VCC = +5 V (pin 13), VEE = -15 V (pin 3). A compensation capacitor of **0.1 uF** goes from pin 16 to VEE. The op-amp (LM741/LF356) runs on +/-15 V, with feedback resistor Rf = 5 kohm (a 10 pF capacitor across Rf if needed). - Latch: a 74LS373/74LS374 holds the data at the output port. #### Port address decoding AFH (8085, I/O write) AFH = 1010 1111. For I/O instructions, the address lines A7-A0 appear on both A15-A8 and AD7-AD0. IOW' is low only during an I/O write, so it already includes IO/M. ```text A7,A5,A3,A2,A1,A0 -----+ A6,A4 --[NOT]----------+--[8-input NAND]-- output L (low only when address = AFH) L and IOW' ---[NOR]---> G (clock) of 74LS373 (latch); G is high only when both are low ``` ```text 8085 AD0-AD7 -> 74LS373 (D0-D7 -> Q0-Q7) -> DAC1408 A8..A1 (MSB = A1) +10 V --[R14=5k]--> pin 14 ; pin 15 --[5k]--> GND pin 4 (Io) --> op-amp (-) ; Rf = 5k between output and (-) op-amp (+) -> GND ; Vout = 0 to +9.96 V pin 16 --[0.1 uF]--> VEE (-15 V) ``` #### Program ```text MVI A,80H ; 80H -> about 5 V OUT AFH ; send to output port AFH HLT ``` To generate 0-10 V in steps, send codes 00H to FFH with OUT AFH. The output voltage for code N is 10 x N / 256 V (resolution 39 mV). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With Vref = +10 V and R14 = 5 kohm, what is Iref? A. 5 mA B. 2 mA C. 10 mA D. 0.5 mA Answer: B. Iref = Vref / R14 = 10 V / 5 kohm = 2 mA. (Key point: Reference current) 2. With Rf = 5 kohm, what is Vout for input code N? A. 10 x N/256 V, so FFH gives 9.96 V B. 5 x N/256 V C. 10 x N/255 V exactly 10 V at FFH only D. 2 x N V Answer: A. Vout = Io x Rf = 2 mA x 5 kohm x N/256 = 10 N/256 V. (Key point: Output voltage) 3. What is the 74LS373 latch for, with the address decoder? A. To generate the clock B. To amplify the DAC output C. Set the reference voltage D. To hold the data at port AFH since the 8085 data is only valid briefly during the write Answer: D. The decoder output with IOW' clocks the latch so the DAC input stays steady after the OUT AFH. (Key point: Latch and decoding) 4. Where is the 0.1 uF compensation capacitor connected on the MC1408? A. Across Rf B. Across the supply C. From pin 16 to VEE D. From pin 4 to ground Answer: C. Pin 16 (compensation) takes a 0.1 uF capacitor to VEE to stabilise the reference amplifier. (Key point: Compensation capacitor) ## Related in Instrumentation II - [Interfacing 10-bit DAC with 8085](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Adac10-8085) — Instrumentation II, chapter 4, asked 2× - [8-bit DAC LSB, MSB, resolution, full scale](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Adac-lsb-msb-resolution) — Instrumentation II, chapter 4 - [10-bit DAC triangular wave at 500 Hz](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Adac10-triangular-500hz) — Instrumentation II, chapter 4 - [8086 RAM/EPROM interfacing with reset address](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Arom-ram-8086-fffffh) — Instrumentation II, chapter 1 - [DAC square wave generation at 1 kHz](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Adac-square-wave-1khz) — Instrumentation II, chapter 4 - [ADC at 80H with 8255A: display on port B LEDs](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Aadc-8255-80h-leds) — Instrumentation II, chapter 2 ## Related in other subjects - [Address decoding for 8 LEDs output port](https://www.bctengineer.com/question.md?key=semester_4%2Fmicroprocessor%3Adecoder-output-port-8-leds) — Microprocessor, chapter 4 - [Memory capacity from range and interfacing](https://www.bctengineer.com/question.md?key=semester_4%2Fmicroprocessor%3Amemory-range-capacity-interfacing) — Microprocessor, chapter 4 - [Address decoding for input port at 8050H](https://www.bctengineer.com/question.md?key=semester_4%2Fmicroprocessor%3Adecoder-input-port-lda-8050h) — Microprocessor, chapter 4 - [Address decoding for input and output ports](https://www.bctengineer.com/question.md?key=semester_4%2Fmicroprocessor%3Adecoder-input-output-port-block-decoder) — Microprocessor, chapter 4